The light rays having photons of energy $4.2\,eV$ are falling on a metal surface having a work function of $2.2\,eV$. The stopping potential of the surface is $.........\,V$.

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : The photoelectric effect does not take place,if the energy of the incident radiation is less than the work function of a metal.
Reason $R$ : Kinetic energy of the photoelectrons is zero,if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements,choose the most appropriate answer from the options given below.

Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

Light of wavelength $3500 \ \mathring A$ is incident on two metals $A$ and $B$. If their work functions are $4.2 \ eV$ and $1.9 \ eV$ respectively,which metal will emit photoelectrons?